Gravitational field of a particle in SR According to special relativity, what is the gravitational field due to a particle moving with a constant velocity v? Would it be correct to assume that the particle has a stronger gravitational field because of the relation $M=\gamma m$, where m is the rest mass and M is the relativistic mass? How is the spacetime affected?
 A: The answer to your question:

Would it be correct to assume that the particle has a stronger gravitational field [...]?

is no, it would not be correct. Here is why. 
Comparing gravitational field  in special relativity to its Newtonian limit means trying to take an ill-defined limit. If one wants to include relativistic corrections such as relativistic expression for the energy, gravitational interaction has to be described by the general relativity. There is a weak field limit of GR,  linearized gravity,  but even in this limit the gravitational field no longer can be described by a single Newtonian potential, so we need a new framework to define the 'strength' of gravitational field. 
If we were talking about moving charge, it would have been wrong to say "the electric field of a moving relativistic charge is weaker  in comparison with the stationary one", the right statement would have been "moving charge has qualitatively different field: the electromagnetic field, which has magnetic component, different dependence on the angles...". 
Such is also situation for the gravitational field of a relativistic mass: It is different from Newtonian limit, not just stronger or weaker. 
To illustrate, let us take the first post-Newtonian approximation for the gravitaional dynamics: Einstein–Infeld–Hoffmann equations which contains $1/c^2$ correction to the point particles accelerations. Taking the equation from Wikipedia page:
$$
\begin{align}
\mathbf{a}_A & = \sum_{B \not = A} \frac{G m_B \mathbf{n}_{BA}}{r_{AB}^2} \\
& {} \quad{} + \frac{1}{c^2} \sum_{B \not = A}
    \frac{G m_B \mathbf{n}_{BA}}{r_{AB}^2}
       \left[ v_A^2+2v_B^2 - 4( \mathbf{v}_A \cdot \mathbf{v}_B) - \frac{3}{2} ( \mathbf{n}_{AB} \cdot \mathbf{v}_B)^2 \right. \\
& {} \qquad {} \left. {} -
        4 \sum_{C \not = A} \frac{G m_C}{r_{AC}} -
          \sum_{C \not = B} \frac{G m_C}{r_{BC}} + \frac{1}{2}( (\mathbf{x}_B-\mathbf{x}_A) \cdot \mathbf{a}_B )  \right] \\
& {}\quad{} + \frac{1}{c^2} \sum_{B \not = A} \frac{G m_B}{r_{AB}^2}\left[\mathbf{n}_{AB}\cdot(4\mathbf{v}_A-3\mathbf{v}_B)\right](\mathbf{v}_A-\mathbf{v}_B) \\
& {} \quad {} + \frac{7}{2c^2} \sum_{B \not = A}{ \frac{G m_B \mathbf{a}_B }{r_{AB}}} + O (c^{-4})
\end{align}
$$
we see a lot of new terms. If we try to test the field of relativistic object by measuring acceleration of a small test particle at rest in our reference frame ($v_A=0$) we will notice that indeed, there is a term proportional to $v_B^2$, which could be interpreted as additional force from added relativistic mass, but we also see its dependence on the angle between $\mathbf{n}$ and $\mathbf{v}_B$, terms proportional to acceleration $\mathbf{a}_B$, $1/r^3$ terms. And this is just the first relativistic correction. 
